SAJC_H1_CHEM_P1_Worked_solutions

1  ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examination (worked solutions) Chemistry Higher 1 8872/01 Paper 1 Multiple Choice 19 September 2016 50 minutes Additional Materials: Multiple Choice Answer Sheet, Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are 30 questions on this paper. Answer a ll questions. For each question there are four possible ans wers A, B, C and D. Choo se the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. This document consists of 13 printed pages including this page.

2  Section A 1 Percentage by mass of C in ethanol = (12 x 2) / [(12 x 2) + 6 + 16] = 0.5217 Thus, Mass of C in 11.5 g ethanol = 0.5217 x 11.5 = 6g Therefore, No. of moles of C in 11.5 g ethanol = 6/12 = 0.5 No. of carbon atoms in 11.5 g ethanol = 0.5 x L = 0.5L or L/2 Answer: B 2   C H Cl % 24.2 4.1 71.2 Divide by Ar 24.2 / 12 = 2 4.1 / 1 = 4.1 71.2 / 35.5 = 2 Divide by smallest number 2 / 2 =1 4.1 / 2 = 2 2 / 2 =1 Ratio 1 2 1 Therefore, empirical formula of CFC-150a is CH2Cl Among the options, only D has this empirical formula Answer: D 3 A: No change in oxidation number (O.N) of S, remaining at +6 B: O.N of S changed from +6 to +4 C: O.N of S changed from +6 to 0 D: O.N of S changed from +6 to -2, thus the biggest difference/change Answer: D

3  4 Since P and Q are in the same period, - P must come after Q to have more protons than Q - To be after Q and have a lower 1st IE, P must be either Gp III (and Q Gp II) or Gp V (and Q Gp VI) as these are where the anomalies of the increasing 1 st IE trend across the period. - If Q is Gp II and P is Gp III, P has one more unpaired electron than Q as the valence electron configuration of P : Q would be ns2 : ns2 np1; If Q is Gp V and P is Gp VI, P has one less unpaired electron than Q as the valence electron configuration of P : Q would be ns2 np4: ns2 np3 Thus, P must be in Gp VI while Q is in Gp V Answer: C 5 A: Ca(NO3)2: There is ionic bonds between Ca 2+ and NO3- as well as covalent bonds within the NO3- anion. B: MgS: There is only ionic bonds between Mg2+ and S2-. C: Only contains metallic bonds given that it’s a metal D: SO2: Only contains covalent bonds given that it’s a giant covalent compound Answer: A 6 Full structure of compound R is   The bond angles present are: - 109.5° around C that have 4 sigma bonds - 180° around the C of CN - 120° around the C of C=C - 105° around the O of O-H Answer: B

4  7 A: False. Its not true as the order should be 4>3>1>2 as 2 is branched while 1 is straight chain so 1 should have more extensive id-id due to larger surface area when compared with 2. B: False. As mentioned abo